Question

A 60 watt filament lamp loses all its energy by radiation from its surface.

The emissivity of the surface is 0.5. The area of the surface is 5 × 109Y m2

.

Find the temperature of the filament (σ = 5.67 × 109[

Jm-2 s

-1 K-4 )​

Answer 1

Given info : A 60 watt filament lamp losses all its Energy by radiation from its surfaces. The emissivity of surface is 0.5. The area of the surface is 5 × 10¯⁵ m².

To find : The temperature of the filament is...

solution : using Stefan's radiation equation, P = σeAT⁴

here σ = Stefan Boltzmann's constant = 5.67 × 10^-8 W/m²k⁴

e = emissivity = 0.5

A = area of surface = 5 × 10¯⁵ m²

P = power = 60 watt

now, 60 = 5.67 × 10^-8 × 0.5 × 5 × 10^-5 × T⁴

⇒60/(5.67 × 2.5) × 10¹³ = T⁴

⇒T⁴ = 4.2328 × 10¹³

⇒T = 2.550 × 10³ = 2550 K

so temperature in degree Celsius is 2550 - 273 = 2277°C

Therefore the temperature of filament is 2277°C (approx).

[note : here answer may not match to your mentioned answer in the book. it's only because of calculation. but you observe , gotten answer is nearest value of your mentioned answer. use approximate value frequently in this type of question. ]

Answer 2

Solution: Given, e = 0.5, A = 5 × 105 m², dQ

dt = 60 W = 60 J s-¹ dQ = eo AT4 dt

We know that

..60=0.5 × 5.67 × 10-8 × 5 × 10-5 × T

. T² = 60×10¹3

5.67×2.5 T = 4.23 × 1013

:.T=(42.3×10¹²) 4

T= 2.55 × 10³ = 2550 K

Explanation:

pwge no,72 physics 12th